The answer is pH = #11.85#.
The balanced chemical equation is
#NaOH_((aq)) + HCl_((aq)) -> NaCl_((aq)) + H_2O_((l))#
SInce #NaOH# is a strong base and #HCl# is a strong acid, they will dissociate completely in aqueous solution; likewise, #NaCl#, being a salt, will dissociate completely in #Na^+# and #Cl^-# ions, thus giving us the complete ionic equation
#Na_((aq))^+ + OH_((aq))^(-) + Cl_((aq))^(-) + H_3^+O_((aq)) -> 2H_2O_((aq)) + Na_((aq))^+ + Cl_((aq))^-#
SInce #Na^+# and #Cl^-# ions are present both on the reactants and on the products' side, they are considered to be spectator ions - which leads us to the net ionic equation
#H_3^+O_((aq)) + OH_((aq))^(-) -> 2H_2O_((l))# - this represents the neutralization reacton between #NaOH# and #HCl#.
You could consider this to be a titration problem - let's assume we are titrating #120.0mL# of #HCL#, #0.0150M#, with #70.0mL# of #NaOH#, #0.0450M#.
We know the volume of #HCl# to be 120.0mL, which translates to a number of moles of
#n_(HCl) = C * V = 0.0150 M * 120.0 * 10^-3 L = 0.0018# moles
Now we add the 70.0mL of #NaOH#. The number of #NaOH# moles added is
#n_(NaOH) = C * V = 0.0450 M * 70.0 * 10^-3 L = 0.00315# moles
Comparing the number of #HCl# and #NaOH# moles, which equal the number of #H_3^+O# and #OH^-# moles, we can see that we have an excess of #OH^-# moles of
#n_(excessOH^-) = n_(OH^-) - n_(H_3^+O) = 0.00315 - 0.0018 = 0.00135# moles left in the solution. This gives us the concentration of #OH^-#
#[OH^-] = n_(excess)/(V_(TOTAL)) = (0.00135 mol es)/((120.0 + 70.0) * 10^-3L) = 0.0071M#
(#V_(TOTAL)# represents the combined volume of the new solution)
So, #pOH = -log([OH^-]) = -log(0.0071) = 2.15#
Therefore, pH #= 14 - pOH = 14 - 2.15 = 11.85#
